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Lab

Lab. “Oreo Lab”. 1. 6.3 Limiting Reagents. Caution: this stuff can be difficult to follow at first. Be patient. Limiting Reagents and Reagents in Excess. So far we have assumed that all of the reactants taking part in the reaction have been transformed into products. (No left-overs)

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Lab

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  1. Lab “Oreo Lab” 1

  2. 6.3 Limiting Reagents Caution: this stuff can be difficult to follow at first. Be patient.

  3. Limiting Reagents and Reagents in Excess • So far we have assumed that all of the reactants taking part in the reaction have been transformed into products. (No left-overs) • Not usually the case. 3

  4. \ Ex. Wood burning • Unlimited oxygen in the air available for the reaction (oxygen = reagent in excess) • When all wood is burned, the reaction stops, no longer using oxygen from the air, and making • no more water or CO2. • The amount of wood limits the amount of products, therefor, it is the limiting reagent 4

  5. Ex. Cake baking • The recipe calls for 2 eggs to combine with 1 cup of flour and this make 1 cake. • How many cakes could you bake if you have 4 eggs in your fridge, and 4 cups of flour in your pantry? • To use up all 4 eggs, you would only need 2 cups of flour. (2 cups left over). • Flour = reagent in excess • Eggs = limiting reagent 5

  6. In Chemistry: Given: 4NH3 + 5O2 6H2O + 4NO Q - How many moles of NO are produced if __ mol NH3 are burned in __ mol O2? a) 4 mol NH3, 5 mol O2 b) 4 mol NH3, 20 mol O2 c) 8 mol NH3, 20 mol O2 • In all of these cases, NH3 limits the production of NO; if there was more NH3, more NO would be produced • Thus, NH3 is called the “limiting reagent” 4 mol NO, works out exactly 4 mol NO, with leftover O2 8 mol NO, with leftover O2

  7. In Chemistry: Given: 4NH3 + 5O2 6H2O + 4NO Q - How many moles of NO are produced if __ mol NH3 are burned in __ mol O2? (d) 4 mol NH3, 2.5 mol O2 • Here, O2 limits the production of NO; if there was more O2, more NO would be produced • Thus, O2 is called the “limiting reagent” • Sometimes the question is more complicated. For example, if grams of the two reactants are given instead of moles. 2 mol NO, leftover NH3

  8. Procedure for solving limiting reagent problems • If not already, balance the chemical equation. • Convert information given into units of moles. • Divide the mole amount of each reactant by its stoichiometric coefficient from the balanced chemical equation. • The reactant with the smallest quotient from step #3 will be the limiting reagent. • Work as typical stoichiometry problem using the limiting reagent.

  9. Example #1 2Ca + O2 → 2CaO How many moles of calcium oxide can you form starting with 15 moles of calcium and 12 moles of oxygen? • If not already, balance the chemical equation. Balanced. • Convert information given into units of moles. Already in units of moles. • Divide the mole amount of each reactant by its stoichiometric ratio from the balanced chemical equation. 15 moles Ca = 7.5 12 moles O2 = 12 2 moles Ca 1 mol O2

  10. Example #1 (con’t) 2Ca + O2 → 2CaO How many moles of calcium oxide can you form starting with 15 moles of calcium and 12 moles of oxygen? • The reactant with the smallest quotient from step #3 will be the limiting reagent.Calcium will be the limiting reagent in this problem. • Work as typical stoichiometry problem using the limiting reagent. 15mol Ca x 2 mol CaO = 15mol CaO 2 mol Ca 10

  11. Example #2 What mass of CO2 can be produced by the reaction of 8.0 grams of CH4 with 48 grams of O2 according to the equation below? CH4 + 2O2 → CO2 + 2H2O • If not already, balance the chemical equation.Balanced. • Convert information given into units of moles. Information given is in units of grams. We need to use the molar mass of each reactant to find moles of each reactant. 8.0g CH4 x 1mol CH4= 0.50mol CH4 16.04g CH4 48g O2 x 1mol O2 = 1.50mol O2 32.0g O2 11

  12. Example #2 (con’t) What mass of CO2 can be produced by the reaction of 8.0 grams of CH4 with 48 grams of O2 according to the equation below? CH4 + 2O2 → CO2 + 2H2O • Divide the mole amount of each reactant by its stoichiometric ratio from the balanced chemical equation. 0.5mol CH4= 0.5 1.5mol O2 = 0.75 1 mol CH4 2mol O2 • The reactant with the smallest quotient from step #3 will be the limiting reagent.CH4

  13. Example #2 (con’t) What mass of CO2 can be produced by the reaction of 8.0 grams of CH4 with 48 grams of O2 according to the equation below? CH4 + 2O2 → CO2 + 2H2O • Work as typical stoichiometry problem using the limiting reagent. 0.5mol CH4 x 1mol CO2 x 44g CO2 = 22g CO2 1mol CH4 1mol CO2

  14. 30.0 g NO = = 30.0 g NO 4 mol NO 4 mol NO 1 mol NH3 1 mol O2 35.3 g NO 22.5 g NO x x x x x x 17.0 g NH3 32.0 g O2 1 mol NO 1 mol NO 4 mol NH3 5 mol O2 Limiting Reagents: shortcut • Do two separate calculations using both given quantities. The smaller answer is correct. Ex3. How many g NO are produced if 20 g NH3 is burned in 30 g O2? 4NH3+5O26H2O+4NO # g NO= 20 g NH3 30 g O2

  15. Assignment • 2Al + 6HCl  2AlCl3 + 3H2 • If 25 g of aluminum was added to 90 g of HCl, what mass of H2 will be produced (try this two ways – with the steps & using the shortcut)? • N2 + 3H2 2NH3: If you have 20 g of N2 and 5.0 g of H2, which is the limiting reagent? • What mass of aluminum oxide is formed when 10.0 g of Al is burned in 20.0 g of O2? • When C3H8 burns in oxygen, CO2 and H2O are produced. If 15.0 g of C3H8 reacts with 60.0 g of O2, how much CO2 is produced? • How can you tell if a question is a limiting reagent question vs. typical stoichiometry?

  16. 2.0 g H2 3 mol H2 1 mol HCl 1 mol Al 1 mol HCl x x x x x 36.5 g HCl 1 mol H2 27.0g Al 6 mol HCl 36.5g HCl #mol Al = 25 g Al = 0.926 mol 2mol 2 = 0.463 # mol HCl= 90 g HCl = 2.466 mol 6mol =0.411 HCl is limiting. # g H2 = 90 g HCl = 2.47 g H2

  17. 2.0 g H2 2.0 g H2 3 mol H2 1 mol Al 3 mol H2 1 mol HCl = 2.47 g H2 = 2.78 g H2 x x x x x x 1 mol H2 1 mol H2 6 mol HCl 2 mol Al 36.5 g HCl 27.0 g Al Question 2: shortcut 2Al + 6HCl  2AlCl3 + 3H2 If 25 g aluminum was added to 90 g HCl,whatmassofH2willbeproduced? 25 g Al 90 g HCl

  18. 17.0 g NH3 17.0 g NH3 2 mol NH3 1 mol N2 2 mol NH3 1 mol H2 = 28.3 g H2 = 24.3 g H2 x x x x x x 1 mol NH3 1 mol NH3 3 mol H2 1 mol N2 2.0 g H2 28.0 g N2 Question 3: shortcut N2 + 3H2 2NH3 If you have 20 g of N2 and 5.0 g of H2, which is the limiting reagent? # g NH3= 20 g N2 # g NH3 = 5.0 g H2 N2 is the limiting reagent

  19. 102.0 g Al2O3 102.0 g Al2O3 2 mol Al2O3 2 mol Al2O3 1 mol Al 1 mol O2 = 42.5 g Al2O3 = 18.9 g Al2O3 x x x x x x 3 mol O2 1 mol H2 1 mol H2 4 mol Al 32.0 g O2 27.0 g Al Question 4: shortcut 4Al + 3O2 2 Al2O3 What mass of aluminum oxide is formed when 10.0 g of Al is burned in 20.0 g of O2? # g Al2O3= 10.0 g Al # g Al2O3= 20.0 g O2

  20. 44.0 g CO2 44.0 g CO2 3 mol CO2 3 mol CO2 1 mol O2 1 mol C3H8 = 49.5 g CO2 = 45.0 g CO2 x x x x x x 1 mol CO2 5 mol O2 1 mol C3H8 1 mol CO2 44.0 g C3H8 32.0 g O2 Question 5: shortcut C3H8 + 5O2 3CO2 + 4H2O When C3H8 burns in oxygen, CO2 and H2O are produced. If 15.0 g of C3H8 reacts with 60.0 g of O2, how much CO2 is produced? # g CO2= 15.0 g C3H8 # g CO2= 60.0 g O2 5. Limiting reagent questions give values for two or more reagents (not just one)

  21. 6.4 Assignment Con’t 34g NH3 19g H2 excess 23.0L H2 25g Mg excess 164.3g Ag3PO4 135.6g Na3PO4 • How many grams of NH3 can be produced from the reaction of 28g of N2 and 25g of H2? • How much of the excess reagent in Problem 6 is left over? • What volume of hydrogen at STP is produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl? • How much of the excess reagent in Problem 3 is left over? • Silver nitrate and sodium phosphate are reacted in equal amounts of 200g each. How many grams of silver phosphate are produced? • How much of the excess reagent in problem 5 is left? 21

  22. 6.4 Assignment 87.9g FeS 13.0L H2 11.9g NaNO3 22.5g H2S 6.45mol Al2O3 • Iron reacts with sulfur to form iron(II) sulfide. You have 32.0g of sulfur and 100g of iron. Calculate the number of grams of iron(II) sulfide that will form. • Zinc reacts with sulfuric acid in a single replacement reaction. You have 40g of zinc and 57g of sulfuric acid. What volume of hydrogen gas will form? • Silver nitrate and sodium bromide will react in a double replacement reaction. You have 24g of silver nitrate and 39g of sodium bromide. How many grams of sodium nitrate will form? • When hydrochloric acid reacts with iron(II) sulfide, a double replacement reaction is to be expected. You have 67g of hydrochloric acid and 58 of iron(II) sulfide. How many grams of hydrogen sulfide will form? • When aluminum is heated in oxygen, aluminum oxide is formed. You have 384g of each reactant. How many moles of aluminum oxide will form? 22

  23. 6.4 Limiting Reagents C.P. • A mixture of 49.1g of tin(II) nitrate and 81.2g of sodium chloride react. How many grams of tin(II) chloride are produced? How much excess is left over? • Silicon monocarbide, commonly known as carborundum, is prepared by heating silicon dioxide in the presence of carbon. How many grams of silicon monocarbide can be formed from 100.0g of graphite and 100.0g of silicon dioxide? (Clue: CO2 is the other product) 38.3g SnCl2 66.73g SiC 23

  24. 6.4 Limiting Reagents C.P. • Silver nitrate and sodium phosphate are reacted in equal amounts of 137g each. How many grams of silver phosphate are produced? • How much of the excess reagent in #1 is left? 38.3g SnCl2 66.73g SiC 24

  25. What volume of hydrogen at STP is produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl?How much of the excess reagent in Problem 3 is left over?

  26. Check Point #2 A mixture of 49.1g of tin(II) nitrate and 81.2g of sodium chloride react. How many grams of tin(II) chloride are produced? How much excess is left over?

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